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A pump on the ground floor of a building...

A pump on the ground floor of a building can pump up water to fill a tank of volume `30 m^3` in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?

Text Solution

Verified by Experts

We are given: volume of water = `30 m^3 `
Time, `t = 15 `minutes = `15 xx 60 = 900 s `
`h= 40 m `
Efficiency, `eta = 30%`.
As density of water = `rho = 103 kg m^(-3)`
Therefore, mass of water pumped, m= volume `xx` density = `30 xx 10^3 kg`.
We know that
Output power `P_0 = W/t = (mgh)/t = (30 xx 10^(3) xx 9.8 xx 40)/(900) = 13070 watt`.
If `P_t` is input power . then
`eta = (P_0)/(P_t) implies P_i = (P_0)/(eta) = (13070)/(30//100) = 43.567 kW`
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